The Phase Rule PDF

-1- The Phase Rule 1.1 Introduction The phase rule is a physical law that regulates the conditions governing equilibrium in a system. This rule finds numerous applications in the chemical engineering field such as petroleum refining, production of chemicals, ceramic in...

-1- The Phase Rule 1.1 Introduction The phase rule is a physical law that regulates the conditions governing equilibrium in a system. This rule finds numerous applications in the chemical engineering field such as petroleum refining, production of chemicals, ceramic industry, glass and cement industries, metallurgical industries, etc… In order to understand the significance and the way of applying this rule, one has first to understand the meaning of some associated terms. 1.2 Terminology 1.2.1 The system A system is a fixed portion of space containing one or more chemical species and having some definite boundary separating them from their surroundings. The nature of a system depends upon the definition of the chemical species it contains. Consider for example a mixture of sand and water placed in a beaker. If we consider the system as consisting of sand and water, then its surroundings will be the beaker and everywhere else beyond this beaker. If the system is defined as being sand, water and beaker then the surroundings will be the portion of space outside the beaker. 1.2.2 Equilibrium By equilibrium is meant the absence of change or even a tendency to change. For example, in the process of esterification, the reaction between the organic acid and alcohol proceeds at a rate that gradually decreases with time until a constant concentration of all species is reached. If the system consisting of reactants + products is left without changing any parameter (Such as temperature, dilution, etc…) then the concentrations will remain constant. These are then termed equilibrium concentrations. If now, some alcohol is added to the equilibrium mixture, than the reaction will be shifted towards the formation of more ester until a new state of equilibrium is reached. 1.2.3 The phase A phase is a homogeneous portion of a system separated from its surroundings by a fixed boundary that can be physically separated from them. For example, a mixture of sand and sugar will appear under a microscope as consisting of grains of each species surrounded by a definite boundary. We are in presence of two phases that can be physically separated by dissolving sugar in water then filtering off the suspension. Fig (1.1) shows this situation. 1  Fig (1.1) Two phase system On the other hand, a solution of sugar in water will consist of one phase since one cannot distinguish, even under microscope, sugar particles. Likewise, a mixture of gases will always show up as one phase. 1.2.4 The component The number of components in a system is the minimum number of chemical species necessary to express the composition of the phases present in that system. If the chemical species present in the system do not react together then the number of components = number of chemical species. For example, in the case of sugar – sand mixture we are in presence of two components. Similarly, in the case of a sugar solution we have two components (water and sugar). The number of components in air is practically 4: oxygen, nitrogen, water vapor and carbon dioxide. In case two or more species can react together the situation is different. Consider for example, the decomposition of calcium carbonate to calcium oxide and carbon dioxide. If we know that at some time, in a crucible, there is calcium oxide powder with evolution of CO2, then definitely there must exist CaCO3 in a state of decomposition. This means that knowledge of the presence of two species implies the existence of the third. Hence the minimum number of species necessary to express the composition of that system is 2 and not 3. In general: If there are R independent chemical reactions that can occur between the species present, then: Number of components = Number of species – R. 1.2.5 The number of degrees of freedom This is the minimum number of independent variables such as temperature, pressure or concentration that must be fixed to completely describe the state of the system Consider for example liquid water. There are an infinite number of combinations of temperature and pressure where water can remain in its liquid state. For instance, water is in liquid state at 25oC and 1 atm or at 60oC and 2 atm, etc… So the number of degrees of freedom = 2. 2 In case of boiling water, this occurs at some definite temperature for each applied pressure. For example, water will boil at 100oC at 1 atm and 200oC at 14 atm. So in order to have boiling, we can fix one variable only (Here pressure) and boiling will take place at some definite temperature. The number of degrees of freedom is therefore 1. 1.3 The Phase Rule 1.3.1 Statement of the rule This rule was first enunciated in 1875 by J.W. Gibbs and is the basis for studying equilibrium in systems of variable complexities. This rule relates the number of degrees of freedom (𝐹) to the number of phases (𝑃) and number of components (𝐶) in the following way: 𝐹 =𝐶−𝑃+2 (1.1) This is given without proof as its proof requires some bases in thermodynamics that students may not have come across. Eventually, the pressure or temperature or composition or any combination thereof may be kept constant. In that case if there are R restrictions on the state of the system, the phase rule takes the form: 𝐹 =𝐶−𝑃+2−𝑅 (1.2) For example, in isobaric systems, pressure is kept constant. The phase rule will read: 𝐹 =𝐶−𝑃+2−1=𝐶−𝑃+1 1.4 Solved examples Example 1.1 Find the number of degrees of freedom in the following situations: A boiling mixture of water and alcohol A boiling mixture of oil and water Decomposition of calcium hydroxide in air A mixture of NaCl, KCl and LiCl in equilibrium with their melt in air Decomposition of HI into iodine and hydrogen Melting in air of an alloy of cobalt and nickel (totally soluble in solid state) Solution: Alcohol and water are totally soluble so: P = 2 (liquid + vapor), C = 2 F=2–2+2=2 Oil and water are totally insoluble so: P = 3 (2liquids + vapor), C = 2 F=2–3+2=1 There is one chemical reaction: Ca(OH)2 = CaO + H2O, so: C = 3 – 1 = 2 P = 3 (Calcium hydroxide, calcium oxide and CO2) and R = 1 (constant pressure): 3  F=2–3+1=0 This is called an invariant situation which must occur at constant temperature. Here, C = 3, P = 4 and pressure is constant (R = 1): F = 3 – 4 + 1 = 0 There is one chemical reaction: HI (g) = I2 (g) + H2 (g), so: C = 3 – 1 = 2 P = 1 (One gaseous phase): F = 2 – 1 + 2 = 3 There is only one solid phase (alloy) and one liquid phase: P = 2, C = 2 and R = 1 (Constant pressure) F=2–2+1=1 Example 1.2 In the blast furnace used to produce pig iron, the following chemical compounds are present: C(s), CaCO3(s), CaO(s), CO2(g), CO(g), O2(g), FeO(s), Fe2O3(s), Fe3O4(s) and Fe(l). Can all these species be present at a time? Solution: We are in presence of 6 solid phases, one gaseous phase and one liquid phase. So, P = 8. The number of chemical species = 10. The following 6 independent equations can be written between these constituents: CaCO3 = CaO + CO2 C + O2 = CO2 2C + O2 = 2CO 3Fe2O3 + CO = 2Fe3O4 + CO2 Fe3O4 + CO = 3FeO + CO2 FeO + CO = Fe + CO2 Any attempt to write any other equation will be useless as it can be deduced from the previous ones. Hence C = 10 – 6 = 4 F=4–8+2=-2 So: all these species cannot exist at a time. 4  -2- One component systems 2.1 Introduction In a one component system: 𝐶 = 1. Hence: 𝐹 = 1 − 𝑃 + 2 = 3 − 𝑃. Since the minimum number of phases is 1, then the maximum number of degrees of freedom is 2. These are temperature and pressure. That is why equilibria in one component systems are described on Pressure – temperature diagrams. Phase transitions in such systems are classified according to whether they are accompanied or not with an observable change in volume. Accordingly, we differentiate between two types: Type I: No appreciable change in volume: Such as melting (or freezing) and allotropic transitions. Type II: there is an appreciable change in volume: Such as vaporization (or condensation) and sublimation (or solid condensation). 2.2 The Clausius – Clapeyron equation The basic equation describing the dependence of vapor pressure on temperature in any phase transition is the Clapeyron equation: 𝑑𝑝0 ∆𝐻𝑣 = (2.1) 𝑑𝑇 𝑇.∆𝑉 Where: 𝑇 is the temperature (K) 𝛥𝑉 is the molar change in volume upon transition from state (1) to state (2) (m3.mol-1) = 𝑉2 − 𝑉1 𝛥𝐻 is the heat effect accompanying the transition (J.mol-1) 𝛥p0 is the differential increase in vapor pressure (Pa) The slope of the po – T diagram is therefore represented by the RHS of this equation. 2.2.1 Class I transitions In class I transitions: 𝛥𝑉 ≈ 0 and the slope = ∞. This means that the po – T line is vertical indicating that the transition temperature is not affected by any change in pressure. As a matter of fact, this is not totally true since most solids increase in volume on melting. For example, solid iron has a density of 7.87 g.cm-3, while molten iron has a density of 6.98 g.cm-3. This means that their specific volumes are: For Fe(s): 0.127 cm3.g-1 and for Fe(l) = 0.143 cm./-1g. Since the atomic mass of Fe = 55.84, their molar volumes are: 𝑉1 = 7.09 cm3.mol-1 and 𝑉2 = 7.98 cm3.mol-1. So that: 5 𝛥𝑉 = 0.89×10-6 m3.mol-1 > 0 This means that the slope of the 𝑃0 /𝑇 melting line will be slightly positive. Water and bismuth represent an exception to that rule. Their volume shrinks upon melting. For example, the specific volume of liquid water is about 18 cm3.mol-1, while it is 19.62 cm3.mol-1 for ice, indicating that 𝛥𝑉 = – 1.62×10-6 m3.mol-1 < 0. These situations are clarified in Fig (2.1). po ΔV=0 ΔV0 T Fig (2.1): Low volume change transitions 2.2.2 Class II transitions In class II transitions, the change in volume is appreciable such as in vaporization. Assuming the vapor to behave as ideal gas, then the molar volume of vapor is related to temperature by the equation: 𝑅. 𝑇 𝑉𝑔 = 𝑝0 𝑅 being the general gas constant = 8.134 J.mol-1.K-1 Neglecting the volume of liquid compared to that of vapor, 𝛥𝑉 ≈ 𝑉, the Clapeyron equation can be rewritten as: 𝑑𝑝0 ∆𝐻𝑣 𝑝0.∆𝐻𝑣 = = (2.1) 𝑑𝑇 𝑇.𝑉𝑔 𝑅.𝑇 2 Assuming ΔHv, the molar latent heat of vaporization to be constant, separating variables and integrating, we get: 𝑑𝑝0 ∆𝐻𝑣 ∫ =∫. 𝑑𝑇 Or: 𝑝0 𝑅.𝑇 2 ∆𝐻𝑣 1 ln 𝑝0 = 𝐴 −. (2.2) 𝑅 𝑇 This is known as the Clausius – Clapeyron equation 6 According to that equation, the vapor pressure will increase exponentially with temperature as can be seen from Figure (2.2). p0 T K Fig (2.2): Effect of temperature on vapor pressure According to equation (2.3), a plot of ln 𝑝0 against 1/𝑇 should yield a straight line ∆𝐻 of slope = − 𝑣. This is used to measure the average latent heat of vaporization of 𝑅 a liquid over the temperature range investigated. It should be noted that from the properties of logarithms any units of pressure can be used. Example 2.1 The following data were obtained on following the variation of vapor pressure of ethanol with temperature over the range 10 – 60oC. Calculate the average latent heat of vaporization of ethanol on this temperature range. 𝑻oC 10 20 30 40 50 60 𝒑0 mmHg 23.6 43.9 78.8 135.3 222.2 352.7 Solution The following table shows the values of ln 𝑝0 against 1/𝑇 (K-1) obtained. 𝟏/𝑻 0.00353 0.00341 0.0033 0.003195 0.0031 0.003 𝟎 𝐥𝐧𝒑 3.16125 3.78191 4.366913 4.907495 5.40358 5.86562 7 The plot of ln p0 against 1/𝑇 is shown in Figure (2.3) 6 ln p0 5.5 5 4.5 4 3.5 3 2.5 2 0.003 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 1/T K-1 Fig (2.3) Plot of ln p against 1/T The slope can be obtained by the mean of any two points lying on the line: ∆𝐻𝑣 5.40358 – 4.907495 Slope = − = = – 5221.94 𝑅 0.0031−0.003195 Considering 𝑅 = 8.314 kJ.mol-1.K-1, we get: ΔHv = 43415 J.mol-1 2.2.3 Prediction of boiling point at different pressures Another form of equation (2.3) can be deduced that helps predict the boiling point of a liquid at any applied pressure. This is done by integrating equation (2.2) between two limits of temperatures 𝑇1 and 𝑇2. This way we obtain: 𝑝20 ∆𝐻𝑣 1 1 ln =.( − ) (2.3) 𝑝10 𝑅 𝑇1 𝑇2 It should be noted that in integrating equation (2.2), the latent heat ∆𝐻𝑣 has been assumed to be constant. If this is not the case, then its relation to temperature must be determined first and included in the integration term. The application of this formula is illustrated in the following example: Example 2.2 Estimate the boiling point of water at a pressure of 10 atm if the latent heat of vaporization = 540 cal.g-1. 8 Solution First, we convert the latent heat to proper units: ∆𝐻𝑣 = 540×18×4.18 = 40629.6 J.mol-1 Hence ∆𝐻𝑣 𝑅 = 4886.9 J.mol-1K-1 The boiling temperature at 1 atm is 373 K. Applying equation (1.4): 10 1 1 ln = 4886.9 × ( − ) 1 373 𝑇2 From which: T2 = 452.5 K ≡ 179.5oC Example 2.3 An empirical equation that is often used to predict the vapor pressure dependence of organic liquids on temperature is the Antoine equation: 𝐵 ln 𝑝0 = 𝐴 − 𝑇+𝐶 Where 𝐴, 𝐵 and 𝐶 are constants. For propanol, these constants are: 𝐴 = 19, 𝐵 = 5000 K −1 and 𝐶 = 40 K when 𝑝0 is expressed in mmHg. Evaluate the latent heat of vaporization at 340 K. Solution: 5000 ln 𝑝0 = 19 − 𝑇 + 40 Differentiating with respect to T, we get: 𝑑𝑝0 5000 0 = 𝑝 𝑑𝑇 (𝑇 + 40)2 𝑑𝑝0 ∆𝐻𝑣 From equation (2.2): = 𝑝0 𝑑𝑇 𝑅.𝑇 2 5000 ∆𝐻𝑣 Hence: (340+40) 2 = 8.314×3402 From which: ΔHv = 33200 J/ mol 2.3 Typical cases 2.3.1 The water diagram 9  Fig (2.4) Water phase diagram This is a typical case of a one – component system as water can exist into three different forms: Solid, liquid and gas. Fig (2.4) Three curves appear on that diagram: The curve BC denotes the relation between vapor pressure and freezing point. It is called the freezing curve. Its slope is negative as the transition from solid to liquid in case of water is accompanied by a decrease in volume. The curve BD shows the relation between vapor pressure and boiling temperature. It is called the vaporization curve. At point D the boundary surface between liquid and vapor vanishes. This is the critical point of water. (374oC).This curve shows the increase in vapor pressure following an increase in boiling temperature as expected from equation (2.3). Under atmospheric pressure (101.3 kPa), a horizontal line crosses the freezing curve at 0oC and the vaporization line at 100oC. The curve AB denotes the transition from solid to vapor phase. This can only occur at very low pressures. This is called the sublimation curve. Inside any of the one phase regions, P = 1 and C = 2. So, pressure and temperature may be varied independently without change in phase. On the other hand, along any of the three curves, they will exist two phases and hence C = 1. This is expected since any of the three curves represents a relation between pressure and temperature and so if one variable is known the other one is automatically fixed. 10 The three curves can be extrapolated to meet at one point. At that point, known as the triple point, the three phases are at equilibrium. In theory, this would mean that under these conditions, one would observe boiling water and ice at a time. The conditions for triple point are: 0.01oC and 611.73 Pa. At such point P = 3 and F = 0 (Invariant point). Normally, it is not easy to reach the triple point since any slight variation in temperature or pressure will get at least one of the three phases to disappear. If water is cooled at an extremely low rate below freezing it may sustain its liquid form in what is known as metastable state. This requires water to be very pure since the presence of any solid impurity will cause rapid crystallization of ice. This means that the metastable portion of the freezing curve extends between the two other curves as shown in Fig (3.5). P0 Liquid Solid Vapor T Fig (3.5) Metastable states 2.3.2 The sulfur diagram The sulfur diagram is more complicated than that of water since it can exist into two main allotropic forms: A low temperature rhombic form and a high temperature monoclinic form. Fig (3.6) shows this system. Under normal atmospheric pressure, rhombic sulfur transforms into a monoclinic form at point A (95.4oC) which on further heating melts at point B (119.2oC). Then transition to the vapor phase occurs at point C (444.6oC). At a pressure exceeding triple point E (at 154oC and 103 atm), rhombic sulfur melts directly without transforming to the monoclinic state. 11  E A B C H Fig (3.6) The sulfur system At very low pressures, we are in presence of two triple points where sublimation of either allotropic form occurs directly without passing through the liquid state: at pressures below 10-5 atm, the rhombic forms sublimate directly. At pressures between 10-5 and 10-4 atm, the rhombic form transforms first to monoclinic which then sublimates to vapor form. At pressures ranging from 10-4 to 103 atm, the transitions are like those under atmospheric pressure. It should be noted that the reason for the presence of the triple point E is that the volume change in the transition from rhombic to monoclinic is less than that associated with melting of the monoclinic form. This is evidenced by the higher slope of the rhombic – monoclinic transition line compared to that of the melting curve of monoclinic form. Accordingly, the two curves can meet at one point. Extremely slow heating of rhombic sulfur may result in “short circuiting” the monoclinic phase. This means that there will be one triple point only where rhombic sulfur, liquid melt and vapor are at equilibrium (Point H). 12  -3- Two-component (Binary) systems 3.1 Introduction In two component systems, C = 2 so that the phase rule reads: F = 4 – P. Since the minimum number of phases = 1, then the maximum number of degrees of freedom is 3. This entails having three coordinates to fully represent a two-component system: Composition of one of the two components, temperature and pressure. In solid – liquid systems it is customary to have the pressure kept constant at one atmosphere, so that F = 3 – P and two variables are sufficient to describe any system: The composition of one of the two components and temperature. However, in liquid – vapor systems the effect of pressure cannot be ignored. Instead of resorting to three dimensional diagrams it is easier to fix either pressure or temperature and study equilibrium on either temperature – composition or pressure – composition diagrams. 3.2 Liquid – vapor systems 3.2.1 Isothermal diagrams In such diagrams the temperature is kept constant, and equilibrium is represented as a relation between pressure and composition (normally molar). The phase rule is written in the form: F = 3 – P. A typical isothermal diagram consists of a plot of total pressure against molar composition of one of the two components A and B. The more volatile component (A) is the one with higher vapor pressure at the specified constant temperature. It is customary to draw the relation between T and xA. In general, if 𝑝𝐴 and 𝑝𝐵 represent the partial pressures of A and B in vapor phase, then the total pressure is: 𝑃 = 𝑝𝐴 + 𝑝𝐵 (3.1) The relation between the partial pressure of each component in vapor phase and its molar composition in liquid state is usually a complex function and its deduction requires special thermodynamic knowledge. This is beyond the scope of this course. However, some special cases arise that can be treated in an easier way. a. Ideal solutions These are solutions that obey Raoult’s law which assumes a direct proportionality between the partial pressure of each component and its mole fraction in liquid phase. This is expressed as follows: 𝑝𝐴 = 𝑝𝐴0. 𝑥𝐴 and 𝑝𝐵 = 𝑝𝐵0. 𝑥𝐵 (3.2) So that: 𝑃 = 𝑝𝐴0. 𝑥𝐴 + 𝑝𝐵0. 𝑥𝐵 (3.3) 13 For an ideal solution to form some necessary conditions must be fulfilled: Complete solubility, similar chemical composition and absence of any tendency for chemical reaction between its components. However, these conditions are not sufficient and such solutions are mainly found in members of homologous series of organic compounds like methanol and ethanol, benzene and toluene, etc. These solutions have zero heat of mixing and there is no change in volume following mixing up the two components. Since 𝑥𝐵 = 1 − 𝑥𝐴 ,, equation (3.3) can be rearranged to read: 𝑃 = (𝑝𝐴0 − 𝑝𝐵0 ). 𝑥𝐴 + 𝑝𝐵0 (3.4) This represents a linear relation between P and xA. This is shown in Fig (3.1) P P = pA + pB 𝒑𝟎𝑨 𝒑𝟎𝑩 pA = 𝒑𝟎𝑨 𝒙𝑨 pB = 𝒑𝟎𝑩 𝒙𝑩 xA B A Fig (3.1): Ideal system A – B The linear relation between the total pressure and composition of A in liquid phase is called the liquidus curve. Now, consider any composition 𝑥𝐴 in liquid phase. At constant temperature, gradual lowering of pressure will cause boiling to start. Let 𝑦𝐴 represent the molar composition of vapor that will be at equilibrium with the liquid of composition 𝑥𝐴. Normally: 𝑦𝐴 > 𝑥𝐴. The relation between 𝑦𝐴 and 𝑥𝐴 is known as the equilibrium relation. In the case of an ideal solution this relation can be easily deduced as follows: According to Raoult’s law: 𝑃 = (𝑝𝐴0 − 𝑝𝐵0 ). 𝑥𝐴 + 𝑝𝐵0 According to Dalton’s law: 𝑦𝐴 = 𝑝𝐴 /𝑃 (3.5) 14  0 𝑝𝐴.𝑥𝐴 Hence: 𝑦𝐴 = 0 −𝑝0 ).𝑥 + 𝑝0 (3.6) (𝑝𝐴 𝐵 𝐴 𝐵 Now we define the relative volatility of A with respect to B as follows: 𝑦𝐴 /𝑥𝐴 𝛼= (3.7) 𝑦𝐵 /𝑥𝐵 0 0 𝑝𝐴.𝑥𝐴 𝑝𝐵.𝑥𝐵 For an ideal solution: 𝑦𝐴 /𝑥𝐴 = /𝑥𝐴 and 𝑦𝐵 /𝑥𝐵 = /𝑥𝐵 𝑃 𝑃 So that equation (3.7) simplifies to: 0 𝑝𝐴 𝛼= 0 (3.8) 𝑝𝐵 Dividing both numerator and denominator of equation (3.6) by 𝑝𝐵0 we get: 𝛼.𝑥𝐴 𝑦𝐴 = (3.9) (1−𝛼).𝑥𝐴 + 1 Equation (3.9) is nonlinear and so, the relation between the total pressure 𝑃 and 𝑦𝐴 will be nonlinear as well. This is called the vaporous curve. The following example illustrates the steps undertaken to draw both liquidus curves and vaporous curves for an ideal solution. Example 3.1 At 63.5oC, the vapor pressures of methanol and ethanol are 750 and 400mmHg respectively. These two liquids form an ideal solution. Draw the isothermal phase diagram ethanol – methanol at that temperature. Solution The liquidus curve can be simply obtained by joining the two vapor pressures by a straight line. As for the vaporous curve, it is customary to deduce it from the equilibrium curve. The value of α = 750 / 400 = 1.875. The following table shows the values of total pressure against values of xA (A = methanol) calculated from equation (3.4) and the values of yA calculated from equation (3.9). xA 0 0.2 0.4 0.6 0.8 1 P 400 470 540 610 680 750 yA 0 0.3191 0.5556 0.7377 0.8824 1 The following figure shows the 𝑃/𝑥𝐴 and 𝑃/𝑦𝐴 plots. 15  800 700 Liquid 600 500 P mmHg 400 300 Vapor 200 100 B 0 A = methanol 0 0.2 0.4 0.6 0.8 1 xA and yA Fig (3.2): Methanol – ethanol system at 63.5oC Example 3.2 The following data represents the equilibrium relation between toluene (A) and benzene (B) at 50oC. Prove that these data conform to the formation of an ideal solution and deduce the value of the relative volatility. Compare with the actual value at 50oC. xA 0 0.15 0.3 0.45 0.6 0.75 0.9 1 yA 0 0.3 0.5 0.65 0.8 0.88 0.95 1 Solution To prove that this is an ideal solution, equation (3.9) must be applicable. To linearize this equation, we write the reciprocal of the two sides: 1 (1−𝛼).𝑥𝐴 + 1 1−∝ 1 = = + 𝑦𝐴 𝛼.𝑥𝐴 ∝ 𝛼.𝑥𝐴 Hence a plot of 1/yA against 1/xA should give a straight line of slope 1/𝛼. The following table shows the calculated values of 1/xA and 1/yA and Fig (3.3) the corresponding plot. A straight line is obtained of slope = 1/𝛼 = 0.413, corresponding to 𝜶 = 𝟐. 𝟒𝟐 At 50oC, the following values can be obtained from any standard reference for the vapor pressures of toluene and benzene. We get: 𝑝𝐴0 = 742 mHg and 𝑝𝐵0 = 303.5 mmHg. Hence 𝛼 = 2.44 16  4 1/yA = 0.413/xA + 0.592 3.5 1/yA 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 6 7 1/xA Fig (3.3): Linearized plot of equilibrium relation in benzene-toluene system b. Non-ideal solutions In non-ideal solutions, Raoult’s law cannot be applied for the whole range of compositions. However, it holds for very large values of xA ≈ 1. For values of xA ≈ 0, there exists a linear relation between 𝑝𝐴 and 𝑥𝐴 known as Henry’s law. It takes the form: 𝑝𝐴 = 𝐻𝐴. 𝑥𝐴 (3.10) Where 𝐻𝐴 is an empirical constant known as Henry’s law constant which is specific to a certain component A. The deviation from ideal behavior can take several forms: If, as shown in Fig (3.4), the partial pressure of A is always higher than that calculated for an ideal solution, then this is called a positive deviation from Raoult’s law. In that figure the dotted straight line represents ideal behavior, and the actual curve is above that line. Also, the sloe of the initial portion of the curve represents Henry’s law constant. In that case: HA > 𝑝𝐴0 On the other hand, if the partial pressure of A is always lower than that calculated for an ideal solution, then this is called a negative deviation from Raoult’s law. In that case: HA < 𝑝𝐴0. This situation is shown in Fig (3.5). 17  Raoult's law Henry's law Henry's law pA Ideal solution Real solution 0 0.2 0.4 0.6 0.8 1 xA Fig (3.4): Positive deviation from Raoult’s law Ideal solution pA Real solution Raoult's law Henry's law 0 0.2 0.4 0.6 0.8 1 xA Fig (3.5): Negative deviation from Raoult’s law The presence of a positive or negative deviation from Raoult’s law will yield a curved rather than a straight line for the liquidus. In that case both liquidus and vaporous lines in Fig (3.1) will be curved. If the deviation from ideality is marked for both components A and B, then both liquidus and vaporous curves may exhibit maximum or minimum value of pressure for positive and negative deviations respectively as illustrated in Fig (3.6) and (3.67) 18  Z Liquid pA Vapor Vaporous curve Liquidus curve 0 0.2 0.4 0.6 0.8 1 xA Fig (3.6): Formation of a positive azeotrope pA Z Vapor Vaporous curve Liquidus curve 0 0.2 0.4 0.6 0.8 1 xA Fig (3.7): Formation of negative azeotrope In case of positive deviation, the observed total pressure is always higher than that of any of the two components. In case of negative deviation, the observed total pressure is always less than that of any of the two components. 19 c. Application: Elevation in boiling point Assume that the boiling point of pure solvent A is T0 at a pressure p0. If a small quantity of non – volatile material is added the pressure will drop to: 𝑝𝐴0 𝑥𝐴 = 𝑝𝐴0 (1 − 𝑥𝐵 ) This will be accompanied by an elevation in boiling point = 𝛥𝑇. Applying equation (2.4): 𝑝𝐴0 (1 − 𝑥𝐵 ) ∆𝐻𝑣 1 1 ln 0 =.( − ) 𝑝𝐴 𝑅 𝑇0 𝑇0 + ∆𝑇 Or: ∆𝐻𝑣 ∆𝑇 ln(1 − 𝑥𝐵 ) =.( ) 𝑅 𝑇0 (𝑇0 + ∆𝑇) Since ΔT x1.If the original mixture is richer in (A) than the azeotropic mixture then the vapor formed will contain less (A) than the liquid (y2 < x2). If now an azeotropic mixture is boiled the liquid will retain its composition until all of it has transformed into vapor. At that point there is an additional restriction that 𝑥 = 𝑦, and so the number of degrees of freedom 𝐹 = 3– 2– 1 = 0. So that at the azeotropic point, boiling occurs isothermally. 22  Vapor T Liquid 0 0.2 0.4 0.6 0.8 1 xA Fig (3.10): Formation of azeotrope with maximum boiling point A similar behavior can be found in systems with maximum boiling point except that the opposite phenomenon occurs. Generally, at the azeotrope two conditions must be fulfilled: 𝜕𝑇 𝜕𝑇 𝑥 = 𝑦 and ( ) = ( ) (3.15) 𝜕𝑥 𝑧 𝜕𝑦 𝑧 The second condition implies that at azeotrope the rate of change of both bubble and dew points is zero. d. Formation of a miscibility gap In some systems, the two components are not totally soluble in each other in the liquid phase. One such system is the phenol – water system. At relatively low temperatures, two layers are formed: A water rich layer and a phenol rich layer. The ratio between their masses at any temperature can be calculated from the lever arm principle. As the temperature is raised the compositions of the two layers get closer until they coincide at some temperature known as the consolute temperature Tc (68oC in the phenol – water system). Above that temperature there is only one phase left, and the two liquids are miscible. Such systems are often associated with the formation of minimum boiling azeotropes as can be seen in Fig (3.11). In this figure, 𝑥1 and 𝑥2 represent the compositions of the two layers in component A. 23  T Vapor One liquid x1 x2 Two liquids B A 0 0.2 0.4 0.6 0.8 1 xA Fig (3.11): Formation of a miscibility gap Another case can occur if the consolute temperature exceeds that of the azeotrope. This is the case in the ether-water system. In that case at boiling there will exist two liquids in equilibrium with the vapor phase and 𝐹 = 0. So, boiling must occur isothermally along a range of compositions as can be seen in Fig (3.12). This situation is known to correspond to the formation of a heterogeneous azeotrope. This figure also shows the corresponding 𝑥– 𝑦 equilibrium diagram where the azeotropic composition is the point of intersection of this curve with a 45o line at which 𝑥 = 𝑦. In another special case, like in the diethylamine – water system, there exists a miscibility gap above a certain consolute temperature so that below that temperature one liquid phase exists (Fig 3.13). This is known as having a minimum consolute temperature. In that case a heterogeneous azeotrope is usually formed. Finally, if the two liquids are totally immiscible in the liquid phase at all temperatures, then the formation of a heterogeneous azeotrope will imply that all compositions will start boiling at the same temperature (azeotropic temperature). The total pressure of vapor will simply consist of the sum of the vapor pressures of the two liquids at any temperature: 𝑃 = 𝑝𝐴0 + 𝑝𝐵0 (3.16) This situation is shown in Fig (3.14) 24 T Vapor Vapor 2 liquids 2 liquids One liquid B A B A Fig (3.13): Heterogeneous azeotrope (case 2) y Vapor 2 liquids B A x Fig (3.12): Heterogeneous azeotrope Fig (3.14): Total insolubility in (case 1) liquid state Example 3.3 A mixture of carbon tetrachloride and silicon tetrachloride in mass ratio 1:1 is kept at 25oC. At that temperature their vapor pressures are 114.9 mmHg and 238.8 mmHg respectively. Assuming an ideal solution to be formed, calculate the total pressure exerted. Solution The more volatile component is silicon tetrachloride (A). Its molecular mass = 170 and the least volatile component is carbon tetrachloride (B). Its molecular mass = 154. 1 170 Hence xA = 1 1 = 0.475 and xB = 0.525 + 170 154 Hence 𝑃 = 0.475 × 238.8 + 0.525 × 114.9 = 𝟏𝟕𝟑. 𝟕 mmHg 25 Example 3.4 The vapor pressure of liquid ethylene at 0oC = 40.6 atm and = 24.8 atm at - 20oC. Estimate the ideal solubility of ethylene in a liquid at 25oC and 1 atm partial pressure. Solution The relation between vapor pressure and temperature takes the form: 𝐵 ln 𝑝0 = 𝐴 − 𝑇 So that: 𝐵 ln 40.6 = 𝐴 − 273 And 𝐵 ln 24.8 = 𝐴 − 253 Solving, we get: 𝐴 = 9.92 and 𝐵 = 1700 Hence at 25oC, the vapor pressure is obtained from: 1700 ln 𝑝0 = 9.92 − 298 0 Hence: 𝑝 = 68 atm Assuming ideal behavior: 𝑝𝐴 = 𝑝𝐴0 𝑥𝐴 and: 1 = 68. 𝑥𝐴 and𝑥𝑨 = 0.0147 Example 3.5 The liquidus curve in a certain binary mixture A – B can be approximated to a second order polynomial. The following data are given: Boiling point of A = 80oC Boiling point of B = 100oC An azeotrope is formed at 0.75 mol fraction A. Find the azeotropic temperature and identify the type of azeotrope formed. Solution 𝑇 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐, where 𝑥 = 𝑥𝐴. At 𝑥 = 0, 𝑇 = 100 → 𝑐 = 100 At 𝑥 = 1, 𝑇 = 80 → 80 = 𝑎 + 𝑏 + 100 𝑎𝑛𝑑 𝑎 + 𝑏 = − 20 (i) 𝑑𝑇 At 𝑥 = 0.75, = 0 so that: 2𝑎 × 0.75 + 𝑏 = 0 (ii) 𝑑𝑥 Solving equations (i) and (ii), we get: 𝑎 = 40, 𝑏 = −60 and so: 𝑇 = 40𝑥 2 − 60𝑥 + 100 At azeotropic point: 𝑥 = 0.75. We get: 𝑻 = 𝟕𝟕. 𝟓oC (minimum boiling azeotrope) 26 Example 3.6 The following data give the vapor pressure dependence of benzene (B)and water (W) on temperature (oC). These two liquids are totally immiscible in liquid state. To C 20 30 40 50 60 70 80 90 100 p0 B mmHg 74 117 178 266 386 549 766 1048 1411 p0 W mmHg 18 32 55 92 148 231 353 527 760 Estimate the temperature and composition of the heterogeneous azeotrope formed. Solution Since the two liquids are totally immiscible then: 𝑃 = 𝑝𝐴0 + 𝑝𝐵0 We calculate the sum of the two vapor pressures as shown in the following table then plot the total pressure against temperature. At a total pressure of 760 mmHg, we get the azeotropic temperature. Fig (3.15) To C 20 30 40 50 60 70 80 90 100 p0 B mmHg 74 117 178 266 386 549 766 1048 1411 p0 W mmHg 18 32 55 92 148 231 353 527 760 Total P 92 149 234 357 534 781 1119 1575 2171 2500 We get from Figure: Tz ≈ 69oC At that temperature: 2250 For benzene: 𝑝0 ≈ 533mmHg 2000 For water: 𝑝0 = 223mmHg 1750 The molar composition of 1500 0 P mmHg 𝑝𝐴 533 azeotrope = 0 +𝑝0 = 1250 𝑝𝐴 𝐵 533+223 Hence xzA = 0.705 1000 (The experimental values are: 750 Tz = 69.2oC and xzA = 0.7) 500 250 0 0 25 50 75 100 125 ToC Fig (3.15): 𝑷/𝑻 curve 27 3.3 Liquid – solid systems 3.3.1 Introduction These systems are usually treated as isobaric systems (𝑃 = 1 atm) as they only involve solid and liquid phases. Hence 𝐹 = 3– 𝑃 and the equilibrium is represented as a relation between melting temperature and composition of one of the two components. It is customary to consider component A as the one with the lower melting point. The main types of equilibria in such systems are discussed in the following sections. 3.3.1 Systems with complete solid solubility (Isomorphous systems) This is the case when the two components are completely soluble in a solid state. Such systems are very rare since some necessary conditions must be fulfilled for them to occur. Among these conditions, we state: Very close atomic or ionic radius Equal valences No tendency for the two components to react together. The fulfillment of these conditions doesn’t necessarily mean that the two components will dissolve in solid state over their whole composition range since these conditions while being necessary are not sufficient. Among such systems we mention the cobalt – nickel system, the FeO – MgO systems. The solubility of A in B can occur on molecular level in two ways: If the crystal lattice of B (host) can accommodate A atoms or ions (guest) in any ratio, an atom (ion) of A in its original site. This is known as substitutional solid solution. This is the case when as complete series of solid solutions is formed. (Fig 3.16). Such solutions can be ordered (Fig 3.16 a) or disordered (Fig 3.16 b). Fig (3.16): a – Ordered substitutional s.s. b – Disordered substitutional s.s. 28 If the host lattice of B can only accept atoms or ions of A to a limited extent, then an intersticial solid solution is formed. In that case the solubility of A in B is usually limited to a few atom%. This is a common occurrence in many systems as the atomic (or ionic) radii of A and B are often very different in values and so the lattice of B will remain stable if only a limited amount of A atoms can squeeze themselves into the interplanar sites of B. This case is shown in Fig (3.17). Fig (3.17): Intersticial solid solution Consider now the system MgO – FeO shown in Fig (3.18). The melting point of MgO is 2825oC and that of FeO is 1371oC. These two oxides dissolve completely in the solid state at all compositions forming one solid solution (s.s.). Let a mixture of the two oxides have an initial composition of, say 70% FeO. As the temperature is raised, melting starts at a certain temperature 𝑇1 that will not remain constant as melting proceeds. This is since on melting we are in presence of two phases: s.s. and liquid so that 𝐹 = 1 and temperature cannot remain unchanged. The locus of initial melting points is termed the solidus curve. Below this curve there will only be one solid phase (s.s.). As heating proceeds, the amount of liquid will increase on expense of solid phase until all solid disappears as we reach a temperature 𝑇2. The locus of final melting points is called the liquidus curve. Above this curve the stable phase is the liquid phase. If one starts with that liquid phase at high temperature and cool down the melt, then a cooling curve can be drawn that shows the variation of temperature with time. This is shown in Fig (3.19) for the chosen composition. At no time will the temperature remain constant throughout the whole cooling operation. Because of the difference in specific heat of liquid and solid phases, the cooling curve will show breaks as the vertical composition line crosses the liquidus and solidus curves. The heating and cooling behaviors explained above will hold for any initial composition. Finally, it is worth mentioning that the relative amounts of liquid and solid phases in the two-phase region can be calculated using the lever arm principle provided the compositions are expressed as mass percent and not mol percent. 29  3500 T oC 3000 2500 Temp oC 2000 s.s. + Liquid 1500 s.s. 1000 500 0 0.2 0.4 0.6 0.8 1 Time % FeO Fig (3.18): MgO – FeO diagram Fig (3.19): Cooling curve 3.3.2 Systems with partial insolubility in solid state As previously mentioned, few systems show total solubility of the two components in solid state. The more common case is that the two components are partially soluble only in solid state. One example is the Zircon oxide ZrO2 (Z) – Thorium oxide ThO2 (T) system shown in Fig (3.20). Liquid 3040 Liq + s.s. 2680 Solid Solution 2600 T(ss) + Z(ss) T Z Fig (3.20): The TiO2 – ZrO2 system 30  In the above figure at temperatures < 2600oC, there exists a miscibility gap where two solid phases coexist: A zircon oxide solid solution where ZrO2 has dissolved some TiO2 and a titanium dioxide solution where TiO2 has dissolved some ZrO2. Above that temperature the two oxides become totally soluble in each other and one solid solution exists. Its melting behavior is then like that explained in Fig (3.18). In most cases, however, the upper temperature of the gap is higher than the solidus temperature. This means that the two solid solutions remain immiscible as melting begins. In that case there will exist three phases at equilibrium: The two solid solutions and liquid. Hence 𝐹 = 0 and melting must start at constant temperature. One such system is the CaO – MgO system shown in mol% in Fig (3.21). T MgO(ss) CaO(ss) Time Fig (3.21) The CaO – MgO system All compositions lying between 21% and 96% MgO start melting isothermally at a fixed temperature of about 2370oC. This is called the eutectic temperature. We must distinguish three types of melting behavior in such systems: Compositions not lying within the range 21 – 96% MgO. For example, the mixture containing 9% MgO shows two solid solution phases at low temperatures up to 1900oC where one CaO solid solution will exist up to 2540oC where melting starts. Melting ends at the liquidus temperature of about 2800oC. Compositions lying within the range 21 – 96% MgO like the composition containing 63% MgO. This composition shows two solid solutions existing 31  below 2360oC. As this temperature is reached, melting starts isothermally until all CaO (s.s.) disappears and we are left with MgO (s.s) + liquid. This temperature will be the initial melting temperature for all compositions lying in the mentioned composition range. It is called the eutectic temperature. As the temperature is raised, final melting will occur at about 2800oC. If a composition corresponding to point E is heated (42% MgO) it will melt isothermally at the eutectic temperature until all solids melt. This is the minimum melting point of the system. If cooling curves are drawn for these three compositions, there will be an isothermal arrest for compositions lying in the range 21 – 96% MgO. As temperature drops to the eutectic value the eutectic reaction takes place by which the remaining liquid solidifies to two solids. This can be written as: Liquid ↔ CaO (s.s) + MgO (s.s.) More generally the eutectic reaction can be written as: Liquid ↔ Two solids This is not strictly speaking a chemical reaction but rather a physical change. It is a reversible reaction so that the two solids will form a liquid. If the composition is less than that of the eutectic all MgO will disappear whereas if it exceeds that of the eutectic all CaO will disappear. Only at the eutectic composition will both solids melt simultaneously. Compositions out of this range will show no isothermal arrests since they do not cross the eutectic line. If one starts with pure CaO at 2200oC and the MgO content is gradually increased, then a CaO solid solution first forms until the horizontal line crosses the boundary between CaO (s.s.) and the two-phase region. As the MgO content is increased, two phases will coexist until about 96% MgO when the second boundary curve is crossed whereby only one phase (MgO s.s.) will be left. These two boundaries are called solvus curves. They represent the extent of solubility of each oxide in the other. Typical systems with partial insolubility in solid state include Titanium oxide – beryllium oxide, aluminum oxide – magnesium aluminate, Fe2O3 – Fe3O4, etc… 3.3.3 Systems with complete insolubility in solid state In such systems the two solids are totally insoluble in solid state so that no solid solutions are even formed. The situation appears as in Fig (3.22) and all compositions will start melting at the eutectic temperature so that the eutectic line is the solidus line of the diagram. Typical examples are NaCl – KCl and Na2O – Fe2O3. 32  Liquid B + Liquid Liquid +A B+A Fig (3.22): Formation of a simple eutectic with no solid solubility Eutectic systems find many practical applications. For example, the NaCl – KCl system plays an important role in heat transfer. NaCl melts at 805oC while KCl melts at 774oC. They form a eutectic at a much lower temperature of 645oC. This means that a eutectic mixture of the two salts can remain in liquid state at temperatures in the 700oC range. Because of the high thermal conductivity of this mixture, it is possible to use it in the liquid state as heat transfer liquid 3.3.4 Systems with partial miscibility in the liquid state It may happen that as solid phases melt, two immiscible liquids are formed. In that case final melting must occur isothermally since F = 0. This defines an invariant point known as the monotectic point. The situation is illustrated in Fig (3.23). Liq 1 Liquid B + Liq1 TB Liq1 + liq2 2 TM M B + Liq2 Liq2 +A B+A x1 x2 B A Fig (3.23): Formation of a liquid miscibility gap 33 In the miscibility gap, there exists two liquids: Liquid1 rich in B and liquid2 poorer in B. As a melt of composition 𝑥1 is cooled, the B rich liquid precipitates solid B as the liquidus curve is crossed. On further cooling the vertical composition line crosses the monotectic line. At that temperature, more solid B is precipitated from the B rich liquid (1) so that below that temperature two phases are left: Solid B + B poorer liquid (2). This means that liquid1 has become poorer in B (liquid2) by precipitating solid B. This defines the monotectic reaction: Liquid (rich in B) = Liquid (poor in B) + B This same reaction takes place for a composition 𝑥2 : On crossing the boundary of the gap, liquid2 splits into two liquid fractions. Then on crossing the monotectic line, the reaction takes place exhausting all liquid1 in favor of forming solid B. This phenomenon is of use in metallurgical and refractory industries since the formation of a monotectic is usually associated with a slight drop in the melting point of B as opposed to its absence where the liquidus drops rapidly on adding A to B (See Figure 3.22). For example, the melting point of SiO2 is about 1720oC. If CaO is added in, say 10% mol, it drops to the monotectic temperature of 1695oC, a drop of 25o. On the other hand, SiO2 doesn’t form a monotectic with Na2O. Addition of 10% Na2O to silica has the liquidus temperature drop to 1400oC, a drop of 320o. 3.3.5 Formation of a congruently melting intermediate compound Sometimes the two components can react to form an intermediate compound AxBy. 𝑥 On molar basis it’s A content will be. On a weight basis the molecular weights 𝑥+𝑦 of A and B must be taken into consideration. When an intermediate compound is formed it may melt congruently or incon- gruently. In the former case, the compound formed retains its composition until it melts to a liquid having the same composition as the original solid compound. In that case the diagram usually consists of two simple eutectic diagrams (as the one shown in Fig 3.22) “sealed” together at a vertical line passing through the composition of the compound formed. Fig (3.24a) shows this situation where the intermediate compound has been labeled C. In that figure it was assumed that there is no solid solubility between the different solid phases. However, this is not always the case. If A or B can dissolve the formed compound C in solid state, we get a diagram as shown in Fig (3.24b) which represents two “sealed” diagrams of the type shown in Fig (3.21). We notice that the composition of C is no longer restricted to a vertical line but rather this phase exists as solid solution over a composition range. This is then known as a non – stoichiometric compound. Such compounds can have different A:B ratios depending on the method of preparation. However, for any of these ratios, they will exhibit the same chemical properties. A common example is FeO which takes the form Fe1-xO with 0.03 < 𝑥 < 0.14. 34  Liquid Liquid Liq +A Liq +A(ss) B(ss) Liq C(ss) B +Liq Liq +C +Liq +C(ss) B +C C+A B(ss) +C(ss) C(ss) + A(ss) B C A B C A (a) (b) Fig (3.24): Formation of a congruently melting intermediate compound (a) Stoichiometric compound (b) Non – stoichiometric compound 3.3.6 Formation of an incongruently melting intermediate compound Sometimes the intermediate compound formed can decompose on melting producing a new solid phase and a liquid. This mode of melting is called incongruent melting since the liquid formed has a composition different from that of the solid compound. Consider Fig (3.25) where the intermediate C decomposes on melting giving off the solid phase B and a liquid of composition designated by point P. This means that on melting three phases will coexist: Solid C, solid B and liquid (P). Consequently 𝐹 = 0, and a new invariant situation arise that defines a reaction known as the peritectic reaction: C(s) = B(s) + Liquid The corresponding temperature is the peritectic temperature while the peritectic point corresponds to the composition of the liquid formed (P). We note that if a melt composition is chosen lying between the composition of C and P, then on cooling there will appear two isothermal arrests: The first as the peritectic reaction occurs and liquid reacts with B to form C and the second being the reaction corresponding to the eutectic formed between B and C (Point E). Compositions having A content less than that of C will not exhibit the eutectic reaction. On the other hand, compositions of A content exceeding that of C will not exhibit the peritectic reaction. In Fig (3.25a) it was assumed that C is a stoichiometric compound. Fig (3.25b) shows the case where there is an extent of solid solubility between the solid phases 35 A, B and C. Examples of systems displaying the above behavior are: BeO – BaO, BaO – La2O3, KPO3 – K4P2O7 etc… Liquid Liquid Liqui B +Liquid B +Liquid B +Liquid d P B(ss) P C +Liq. Liq. + A CC +Liquid +Liquid Liquid Liquid++AA B(ss) + E BC(ss) +C E B+C C(ss)+ A(ss) C+A C+A B C A B C A (a) (b) Fig (3.25): Formation of an incongruently melting intermediate compound (a) Stoichiometric compound (b) Non – stoichiometric compound Finally, it may happen that the intermediate compound C forms a complete series of solid solutions with A. In that case C is a solid solution that decomposes on melting through a peritectic reaction (Fig 3.26). A typical example is the MnO – FeO system. Fig (3.26): Formation of a solid solution that decomposes on melting 36 3.3.7 Subsolidus reactions By subsolidus reactions is meant the transformations occurring in the solid state that don’t involve the formation of any liquid phase. If an intermediate compound is unstable below a certain temperature, 𝑇 ∗ , then it will decompose on cooling to two solid phases. This is called a eutectoid reaction: C(s) = A(s) + B(s) on cooling (Fig 3.27a) On the other hand, if an intermediate compound is unstable above a certain temperature 𝑇 ∗ then it will decompose on heating to two solid phases. This is called a peritectoid reaction. It takes the form: C(s) = A(s) + B(s) on heating (Fig 3.27b) T T B+A B+C C+A T* B+A B+C C+A B C A B C A Fig (3.27a): Formation of a eutectoid Fig (3.27b): Formation of a peritectoid In case solid solutions are formed the diagrams take the shapes shown in Fig (3.28) T T B+A B + C(ss) C(ss)+ A B+A B + C(ss) C(ss) + A B C A B C A (a) (b) Fig (3.28): Subsolidus reactions for solid solutions (a) Formation of a eutectoid (b) Formation of a peritectoid 37 3.3.8 Allotropic transitions in two component systems When one of the components (Say A) undergoes an allotropic transition at a temperature Tt: A(I) = A(II) then the transition will occur isothermally at the same temperature Tt of transition of pure A. Fig (3.29) Liquid B + Liquid Liquid +A B + A (I) T* B + A (II) Fig (3.29): Allotropic transition with no solid solubility If A can dissolve B the situation is different since now A will be in the form of a solid solution including component B. Hence at allotropic transition we are in presence of two phases: A I (ss) and A II (ss) so that 𝐹 = 1 and the transition cannot occur isothermally. Two cases are possible: If the presence of B dissolved in A decreases the transition temperature, then a eutectoid is formed corresponding to the reaction: A I (ss) = A II (ss) + B on cooling (Fig 3.30) And if the presence of B dissolved in A increases the transition temperature, then a peritectoid is formed corresponding to the reaction: A I (ss) = A II (ss) + B on heating (Fig 3.31) 38  Liquid Liquid B + Liquid Liquid + A(I) B + Liquid Liquid + A(I) B A (I) B B + A(I) A (I) B + A(I) A(I) + A(II) A(I) + A(II) A (II) B + A(II) B + A(II) A (II) B A B A Fig (3.30) Allotropic transition Fig (3.31) Allotropic transition with eutectoid formation with peritectoid formation 3.4 Solid – liquid binary systems of industrial importance 3.4.1 The carbon – iron diagram This diagram is one of the most important diagrams of use in the iron and steel industry as well as the use of different iron alloys. Fig (3.32) The presence of carbon in iron results in the formation of one stable compound of composition Fe3C, known as cementite. This compound contains about 6.67% carbon by weight. Pure iron exists in three allotropic forms: α – iron, γ – iron and δ – iron. Their transition temperatures are: α – iron ↔ γ – iron 908oC γ – iron ↔ δ – iron 1400oC It finally melts at 1539oC. As carbon is dissolved in iron as solid solutions, a eutectoid is formed at 723oC and peritectic melting occurs at 1493oC. The α – iron solid solution is known as ferrite and the γ – iron solid solution as austenite. The eutectoid reaction (on cooling) is: γ – iron (austenite) → α – iron (ferrite) + Fe3C (cementite) at 723oC (Point J) The peritectic melting reaction (on heating) is: γ – iron (austenite) → δ – iron + liquid at 1493oC. A eutectic is formed between iron and cementite at 1130oC and about 4.3% carbon (Point E). This is the minimum melting point in that system. 39  Fig (3.32) Carbon – iron diagram In that system, steels are formed at carbon content < 2%. They are classified as hypoeutectoid steels (C < 0.8%) and hypereutectoid steels (C > 0.8%). Cast irons, on the other hand, possess carbon contents > 2% and are classified as hypoeutectic cast irons (2% < C < 4.3%) and hypereutectic cast irons (C > 4.3%). Finally, wrought iron is manufactured at carbon content < 0.022% (Point H). 3.4.2 The Al2O3 – SiO2 diagram This system is of major importance in the manufacture of ceramic products. Fig (3.33). This is since the raw materials of conventional ceramics consist of clays and quartz. Since clays are aluminum silicates, and quartz consists of silica, one can understand the importance of studying that system. 40  Fig (3.33) Silica – alumina diagram Pure silica melts at about 1720oC and pure alumina at 2045oC. One intermediate non – stoichiometric compound is formed in that system: Mullite (3Al2O3.2SiO2) containing about 72% alumina by weight. This compound melts congruently at about 1860oC. It forms a eutectic with silica at about 6% alumina at 1590oC. Mullite also forms a eutectic with alumina at about 77% alumina at 1830oC. This compound has excellent chemical resistance, and its formation is associated with an appreciable increase in mechanical strength. One important result that can be deduced from that system is the drop in melting temperature of silica as small amounts of alumina are added while on adding greater amounts of alumina the melting point steadily increases. This is the basis of classifying aluminosilicate refractory materials. It is worth mentioning that other alumina – silica compounds occur naturally (Like sillimanite Al2O3.SiO2). These are however non – equilibrium compounds that cannot exist on an equilibrium diagram. 3.4.3 The Na2O – SiO2 diagram Fig (3.34) shows the high silica corner of the Na2O – SiO2 diagram. Three intermediate compounds are formed: Na2Si2O5 (Na2O.2SiO2) that melts congruently at about 874oC 41  Na2SiO3 (Na2O.SiO2) that melts congruently at about 1088oC Na4SiO4 (2Na2O.SiO2) that melts incongruently at about 1114oC, dissociating into a liquid of about 40% silica and solid Na2O. There are four invariant situations: A eutectic occurs at about 790oC and 75% silica. The eutectic reaction is: Liquid → Na2Si2O5 + SiO2 A eutectic occurs at about 830oC and 62% silica. The eutectic reaction is: Liquid → Na2SiO3 + Na2Si2O5 A eutectic occurs at about 1015oC and 43% silica. The eutectic reaction is: Liquid → Na2Si2O5 + Na4SiO4 A peritectic occurs at about 1114oC and about 40% silica. The peritectic reaction is: Na4SiO4 → Na2O + liquid (on heating) Fig (3.34) Na2O – SiO2 diagram This diagram is made of use in the manufacture of glass and glazes. The main feature is the serious drop in temperature as sodium compounds are added to silica. For 42 example, in the glass industry the vapors rising in the glass melting furnace are rich in Na+ ions. The furnace roofs are usually made of silica refractory bricks. These vapors can therefore attack the refractory lining forming the eutectic at 790oC. This commonly causes failure of the bricks by partial melting. The resistance of bricks can be improved by proper design of the furnace roof and by proper routing of flue gases. 3.4.4 The CaO – SiO2 system Fig (3.35) shows the high silica corner of the SiO2 – CaO system. One compound only is shown: CaSiO3 (CaO.SiO2 Wollastonite) that melts congruently at 1544oC. This compound forms a eutectic with SiO2 at 63.5% SiO2 and 1444oC. The main feature is the formation of a monotectic reaction at 1679oC owing to the presence of a miscibility gap. This situation is responsible for the negligible drop in temperature (From 1723 to 1679oC) on adding lime (CaO) to silica. This finds its application in the refractory industry as lime is added to increase the rate of reactions in the production of refractory bodies based on silica. This addition has the advantage of unaffecting the melting point of silica. Fig (3.35) CaO – SiO2 diagram 43  Three component (ternary) systems 4.1 Graphical representation 4.1.1 Introduction On treating three component systems, we must use three dimensional diagrams. This is since C = 3, hence; F = 4 – P. Since the minimum number of phases is P = 1, then the maximum number of degrees of freedom = 3. Two of these three are the compositions of two of the three components. The third variable is usually temperature as such diagrams are commonly investigated under atmospheric conditions. Temperature – composition diagrams are usually drawn in form of a right prism the basis of which is used to represent compositions while the vertical edges represent temperature. The compositions are usually plotted on an equilateral triangle. 4.1.2 Composition triangle This is an equilateral triangle at the vertices of which are located the three components (A, B and C) (Fig 4.1). Two rules held in this triangle related to the compositions of the three components: If a straight line is drawn parallel to an edge (say BC) all compositions lying on that line have constant percentage of the opposite edge (A). If a line is drawn through any vertex (say A) all compositions lying on that line will have a constant ratio of the two other components (B:C) A Constant B:C Constant %A B C Fig (4.1): Composition triangle To represent a ternary composition on that diagram one needs to know the compositions of two components only. Say, A and B. So, assume that we wish to locate a composition of 30% A and 50% B on that diagram. We draw a line parallel 44 to BC at 30% A and another line parallel to AC at 50% B. Their intersection is shown as a bold circle. We note that this location corresponds to 20%C. A %B %A 50%B 30%A B C %C Fig (4.2): Locating points on the ternary composition triangle 4.1.3 Temperature – composition Three – dimensional representation As previously stated, the full determination of the state of a ternary system at constant pressure requires knowing its composition and temperature. This is represented by a triangular right prism as shown in Fig (4.3). Once the composition is located on the base triangle as explained above, a vertical line is drawn to reach the required temperature. In that three-dimensional diagram, any horizontal plane will denote an isothermal condition while vertical planes parallel to a face of the prism will indicate constant composition of the component whose vertex is opposed to that face. Also, vertical; planes passing through an edge will denote constant ratio of the two other components. (See the two rules in section 4.1.2). 45  T Isothermal plane A B C Fig (4.3): Temperature – composition diagram 4.2 Liquid – vapor systems 4.2.1 Introduction On dealing with three component liquid – vapor systems, one must consider the relation between composition and boiling point. Since the pressure is kept constant then 𝐹 = 4 − 𝑃 + 1 = 3 − 𝑃. On boiling, 𝑃 = 2 and therefore 𝐹 = 1. So, boiling will not occur at constant temperature but there will rather be a bubble point and a dew point in the same way as in binary systems. This situation is illustrated in Fig (4.4) where the liquidus and vaporous curves of the three binaries: A – B, B – C and C – A are drawn on each edge. The union of the vaporous curves forms as vaporous surface in space. Similarly, a liquidus surface is formed by the three liquidus curves on the edges. Above the vaporous surface the prevailing phase is vapor while liquid phase will prevail at temperatures lower than those of the liquidus surface. 46  TB L1 L2 V1 TA V2 TC A B C Fig (4.4): Liquid – vapor three components system A horizontal isothermal plane drawn will intercept the vaporous surface in a curve like V1V2 and the liquidus surface in L1L2. It is common to use isothermal sections to illustrate equilibria in such systems rather than drawing the three-dimensional diagram. For example, the isothermal section in Fig (4.4) can be drawn to show the two dotted intersections of the horizontal plane with the vaporous and liquidus surfaces. Such a section is shown in Fig (4.5). The region between L1L2 and vertex B consists of a liquid phase since it lies below the liquidus surface while that between V1V2 and the side AC consists of a vapor phase since it lies above the vaporous surface. In between those two curves is a two-phase region. The compositions of liquid and vapor coexisting in equilibrium are joined by tie lines. 47 Consider a liquid composition at room temperature denoted by point X. If heated, it will start boiling when the vertical composition line crosses the liquidus surface. In the two-phase region in the isothermal section shown a tie line can be passed through point X so that the liquid composition will be that corresponding to point l while the composition of vapor present will be that of point v. The relative ratio of these two phases is obtained from the lever arm principle: 𝐿 𝑋𝑣 = (4.1) 𝑉 𝑋𝑙 The details of how to pass a tie line through a given composition will be discussed in the next section. B Liquid L1 L2 X V2 V1 v Vapor A C Fig (4.5): Isothermal section in ternary diagram 4.2.2 Ideal solutions Consider an ideal solution of composition (𝑥𝐴 , 𝑥𝐵 , 𝑥𝐶 ). At any temperature 𝑇, the partial pressure of any component is obtained from Raoult’s law: 𝑝𝐴 = 𝑝𝐴0. 𝑥𝐴 𝑝𝐵 = 𝑝𝐵0. 𝑥𝐵 and 𝑝𝐶 = 𝑝𝐶0. 𝑥𝐶 If the total pressure is P, then: 𝑝𝐴0. 𝑥𝐴 + 𝑝𝐵0. 𝑥𝐵 + 𝑝𝐶0. 𝑥𝐶 = P (4.2) Since each vapor pressure is a function of temperature then the LHS will be a function of temperature as well. A curve is drawn between the LHS and temperature and the bubble point obtained at a total pressure = 𝑃. 48 The dew point is obtained by applying simultaneously Raoult’s law and Dalton’s law: 𝑝𝐴 = 𝑝𝐴0. 𝑥𝐴 = 𝑦𝐴. 𝑃 So that the liquid composition of any component (like A) is: 𝑃 𝑥𝐴 =. 𝑦𝐴 (4.3) 𝑃𝐴0 For a given vapor composition (𝑦𝐴 , 𝑦𝐵 , 𝑦𝐶 ) at any temperature, one can get the values of the vapor pressures and hence the values of the liquid phase molar compositions: (𝑥𝐴 , 𝑥𝐵 , 𝑥𝐶 ). The sum 𝑥𝐴 + 𝑥𝐵 + 𝑥𝐶. is plotted against temperature and the dew point obtained as corresponding to a sum =1. If the corresponding values of (𝑥𝐴 , 𝑥𝐵 , 𝑥𝐶 ). and (𝑦𝐴 , 𝑦𝐵 , 𝑦𝐶 ) are known, then a tie line joining these compositions can be drawn. The equation of the liquidus curve at any temperature is a linear relation obtained from the set: 𝑥𝐴 + 𝑥𝐵 + 𝑥𝐶 = 1 and 𝑝𝐴0. 𝑥𝐴 + 𝑝𝐵0. 𝑥𝐵 + 𝑝𝐶0. 𝑥𝐶 = 𝑃 The equation of the corresponding vaporous curve is not linear and is obtained by passing tie lines through some chosen liquid compositions on the liquidus curve and calculating the corresponding vapor compositions. These calculations are illustrated in the following example. Example 4.1 The following table shows the vapor pressure – temperature data for three liquids A, B, and C forming an ideal solution. For a mixture consisting of 20%A, 30%B and 50%C (by mol) evaluate the bubble point, the dew point under atmospheric conditions and draw an isothermal section at 60oC. Also deduce the composition that will have a bubble point of 65oC and a dew point of 90oC. Temp oC 30 40 50 60 70 80 90 𝒑𝟎𝑨 mmHg 59.4 85.6 120.6 166.4 225.4 300.2 393.8 𝒑𝟎𝑩 mmHg 148.9 212.4 315 443 610.8 826.9 1101 𝒑𝟎𝑪 mmHg 726.1 800 874.8 952.2 1031.3 1172 1194 Solution (1) P = 0.2𝑝𝐴0 + 0.3𝑝𝐵0 + 0.5𝑝𝐶0 The table shows P values at different temperatures Temp oC 30 40 50 60 70 80 90 𝑷 mmHg 419.6 480.84 556.02 642.28 743.97 894.11 1006.1 49 Fig (4.6) shows the plot between P and T 1100 Σp 1000 900 800 700 600 500 400 300 0 10 20 30 40 50 60 70 80 90 100 ToC Fig (4.6): Determination of bubble point From that figure: Bubble point ≈ 71oC (2) To get the dew point, the values of 𝑦𝐴 , 𝑦𝐵 , 𝑦𝐶 are calculated from Dalton’s and Raoult’s laws (Equation 4.3). Their values are shown in the following table Temp oC 30 40 50 60 70 80 90 Σy 4.6135 3.3241 2.4186 1.8272 1.4161 1.1063 0.9113 The plot is shown in Fig (4.7) The dew point is ≈ 86oC. (3) Let a solution composition (𝑎, 𝑏, 𝑐) have a bubble point of 65oC and a dew point of 90oC. The corresponding vapor pressures at both temperatures are interpolated from the given table and shown below: Component 65oC 90oC A 195 393.8 B 567 1101 C 984 1194 50  5 Σy 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 20 30 40 50 60 70 80 90 100 ToC Fig (4.7): Determination of dew point The following equations describe the previous rules: 𝑎+𝑏+𝑐 =1 (i) 195𝑎 + 567𝑏 + 984𝑐 = 760 (ii) 760 760 760 𝑎+ 𝑏+ 𝑐=1 (iii) 394 1101 1194 Solving we get: 𝒂 = 𝟎. 𝟐𝟖𝟏, 𝒃 = 𝟎. 𝟎𝟎𝟓 , 𝒄 = 𝟎. 𝟔𝟑𝟔 (4) Finally, to draw an isothermal section at 60oC, consider any composition: (𝑥𝐴 , 𝑥𝐵 , 𝑥𝐶 ). The equation of the liquidus curve is obtained by considering the two equations: 𝑥𝐴 + 𝑥𝐵 + 𝑥𝐶 = 1 (iv) 0 0 0 𝑝𝐴 , 𝑥𝐴 + 𝑝𝐵 , 𝑥𝐵 + 𝑝𝐶 , 𝑥𝐶 = 760 (v) Substituting with vapor pressure values at 60oC: 166.4𝑥𝐴 + 443𝑥𝐵 + 952.2 = 760 (vi) To plot the liquidus curve we choose for example the BC edge where 𝑥𝐴 = 0. The two above equations simplify to: 𝑥𝐵 + 𝑥𝐶 = 1 443𝑥𝐵 + 952.2 xC = 760 Solving we get: 𝑥𝐵 = 0.377 and 𝑥𝐶 = 0.623. This is shown as point L1 in Fig (4.8). 51 Next set 𝑥𝐵 = 0 so that the two equations become: 𝑥𝐴 + 𝑥𝐶 = 1 166.4𝑥𝐴 + 952.2𝑥𝐶 = 760 Solving we get: 𝑥𝐴 = 0.244 and 𝑥𝐶 = 0.756. This is shown as point L2 in Fig (4.8). Note that this means that the liquidus line crosses the sides BC and AC and not AB. Had we set 𝑥𝐶 = 0 we would have obtained negative values of 𝑥𝐴 and 𝑥𝐵. To get the corresponding vapor compositions, we apply combined Raoult’s and Dalton’s laws: For A component: 𝑝𝐴0. 𝑥𝐴 = 𝑦𝐴. 𝑃, hence on the edge AC: 166.4 × 0.244 = 𝑦𝐴 × 760 which gives: 𝑦𝐴 = 0.0534 and hence 𝑦𝐶 = 0.947 Similarly for B component: 𝑝𝐵0. 𝑥𝐵 = 𝑦𝐵. 𝑃, hence on the edge BC: 443 × 0.377 = 𝑦𝐵 × 760 which gives: 𝑦𝐵 = 0.22 and hence 𝑦𝐶 = 0.78 The two vapor compositions corresponding to liquid compositions L1 and L2 are labeled V1 and V2 on the figure. Although the vaporous curve is nonlinear the compositions are so close that a straight line was drawn to join them. A Liquid L2 V2 Vapor B L1 V1 C Fig (4.8): Isothermal section at 60oC 52 4.3 Solid – Liquid systems 4.3.1 Introduction Three-components solid – liquid systems are usually studied under atmospheric pressure, so that the phase rule can be written in the form: 𝐹 =3−𝑃+2−1=4−𝑃 Since the minimum number of phases is 1, therefore the maximum number of degrees of freedom = 3. This entails the use of three-dimensional representations like those used in liquid – vapor systems (Figures 4.1 to 4.3). 4.3.2 Systems with complete solid solubility This is the case when the three components are totally soluble in the solid state. This situation is extremely rare and of theoretical importance only. This is since it is almost impossible for the conditions previously stated for complete solid solubility (Section 3.3.1) to be fulfilled in three components. Eventually, the ternary diagram will appear as follows: (Fig 4.9) Liquid TB TA TC Solid solution A B C Fig (4.9): Complete solubility in the solid state 53 4.3.3 Systems with complete solid insolubility (a) Three-dimensional representation If the three components are totally insoluble in each other, and no intermediate compounds are formed, then there will exist three binary eutectics: e1 between B and C, e2 between C and A and e3 between A and B. In Figure (4.10) an example of such systems is given: In that case, the melting points of A, B and C are respectively 1190oC. 1080oC and 1160oC. Three binary eutectics are formed: A eutectic between B and C at about 770oC A eutectic between C and A at about 700oC A eutectic between A and B at about 840oC There are also three curved surfaces (shaded) at the three vertices. Each two of these intersect in a curved line beginning at the binary eutectic. All three lines meet at one point, which is the ternary eutectic (At about 680oC). This is the lowest melting temperature of the system. Fig (3.10): Formation of a ternary eutectic 54 The three shaded surfaces represent the liquidus surface of the diagram, while a horizontal plane passing through the ternary eutectic point is the solidus plane of the system. The three lines that join the binary eutectic points to the ternary eutectic represent the boundaries between the three liquidus surfaces and are called boundary lines. The regions in space below the curved surfaces are called primary fields of crystallization. For example, in Fig (4.10), the region below the curves surface at B is called the primary field of crystallization of B, since the first solid that will precipitate from a melt cooled in that region is B. (b) Variation of liquid composition on cooling of a melt Consider, in Fig (4.11) a melt of composition M about component A at a temperature much higher than the liquidus temperature. As it is cooled down, the vertical line denoting its composition intercepts the curved surface at A at point X. Below the temperature corresponding to point X, solid A starts to form. If a straight line drawn in a horizontal plane joining the A-axis to X is extended to meet the curved surface at L, then point L represents the composition of the first liquid formed. As the solid – liquid mixture (A + liquid) is further cooled, the liquid composition follows the dotted arrow starting from point L until it meets the boundary line between A and B at point L’. Below the temperature corresponding to point L’, B starts to precipitate so that three phases are present: A(s), B(s) and liquid. Following the phase rule, 𝐹 = 4 − 3 = 1, and the temperature cannot remain constant, but rather decreases as the liquid composition follows the eE ̅̅̅ boundary line. Finally, as the temperature reaches that of the ternary eutectic E, the following reaction takes place: Liquid = A(s) + B(S) + C(s) This is the ternary eutectic reaction which involves 4 phases keeping the temperature constant as all the remaining liquid solidifies. Below the eutectic temperature, the mixture of composition X will consist of the three solid phases A, B and C. Since the use of three-dimensional representation is not practical, a two-dimensional projection is rather used. On this projection, the sequence of phases appearing on cooling the melt of composition X can be easily followed, the sequence of phases being as follows: Liquid → Liquid + A → Liquid + A + B → Eutectic reaction → A + B + C The variation of the liquid composition represented by the path is called the crystallization path of the mixture (Fig 4.12). 55  M X* X L e L’ A E B Fig (4.11): Cooling behavior of a melt A A L e L’ E B C B C Fig (4.12): Crystallization path of a melt 56 4.3.4 Formation of a congruently melting binary intermediate compound (a) Formation of a true binary Consider a three-component system A – B – C where A and B can react to form a congruently melting compound D. As can be seen in Fig (4.13), the composition of D lies on the edge A – B of the projected triangle ABC. On that edge is drawn the binary A – B diagram showing the formation of two eutectic point: A eutectic between A and D of composition e3 and another one between D and B of composition e′3. If a line is drawn to join C to D, the original triangle will be divided into two triangles, known as tie triangles, and the line AD is a tie line. If C and D form a simple eutectic (e), then the formation of the intermediate compound simply gives way to two triangles like the ones in Figure (4.13) joined by the tie line CD, and therefore we are in presence of two ternary eutectic points E1 and E2. This three-dimensional situation, shown in Fig (4.14), is known as formation of a true binary. The projected diagram shows as in Figure (4.15). e E2 E1 C B D A Fig (4.14): 3-Dim representation of the formation of a true binary 57  A e31 A D D E1 e32 e e2 E2 C B B C e1 Fig (4.15): Formation of a true binary In Fig (4.15), the directions of arrows on boundary lines indicate the direction taken by the liquid compositions as melts are cooled down. All melts (Except for those compositions lying on the tie line CD) ultimately solidify to three solids through the ternary eutectic reactions: Either A, C and D or B, C and D. Fig (4.15) can be used to set some useful rules when dealing with ternary solid – liquid systems: Rule 1: The first solid to precipitate when a melt of a certain composition is cooled down is indicated by the field of crystallization where the composition of the melt lies. For example, if a melt whose composition lies in the zone delimitated by the lines Be1 − e1 E2 − E2 e32 − B, is cooled, the first solid to precipitate is B, since the composition of the melt lies within the primary field of crystallization of B. Rule 2: A ternary invariant point is a eutectic if its composition lies within the tie triangle formed by the same phases surrounding this point. 58 The ternary invariant point E1 lies in the tie triangle ACD and is the intersection of the boundary curves of the phases A, C and D. Since these phases coincide with the vertices of the tie triangle in which the composition of the invariant point lies, then this is a eutectic point. Rule 3: An intermediate compound is congruently melting if its composition lies within its primary field of crystallization. In Fig (4.15), the composition of the intermediate compound D lies within its primary field of crystallization. Therefore, this compound melts congruently. Rule 4: The final solid phases present as a melt solidifies are the vertices of the tie triangle in which the composition of the melt lies. For example, a melt whose composition lies in the tie triangle ACD will solidify completely to the phases A, C and D, through the eutectic reaction: Liquid = B + C + D. The final composition of the solidified mix is that of the original composition of the melt. (b) Formation of a quasi-binary In some instances, an intermediate congruently melting compound can behave in a different manner. Consider Fig (4.16) where an intermediate compound D is formed. This compound lies within its primary field of crystallization so that Rule 3 applies. However, the ternary invariant point R is not a eutectic point as it violates Rule 2. This point is the intersection of the boundary lines between the primary fields of crystallization of the compounds A, C and D, while it lies in the tie triangle BCD. Such an invariant point is called reaction point and does not represent the lowest melting point of the whole diagram. The lowest melting point is that of the ternary eutectic point E. The composition of point E lies in the tie triangle BCD and is surrounded on the diagram by these same three pha